LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every ordered field is an ordered ring, and its order is the one its positive cone induces

Statement

Let $F$ be an ordered field with positive cone $P$ (Ordered field), and let $\le$ be the relation $a \le b :\iff (b - a \in P$ or $a = b)$ that Ordered field defines from $P$ . Then:

$F$ with the operations of Field is a commutative ring (Every field is a commutative ring with $1 \ne 0$ ; it is an integral domain, and it is a commutative division ring), and $P$ is a cone in the sense of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = \{\, x : 0 < x \,\}$ satisfies trichotomy and closure, and $a < b :\iff b - a \in P$ recovers the order;
$\le$ is a total order making $F$ an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), whose positive cone $\{\, x \in F : 0 < x \,\}$ is exactly $P$ ;
$1 \in P$ , that is $0 < 1$ .

So an ordered field is an ordered ring, and its order and its positive cone determine each other exactly as they do in any ordered ring.

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and $\le$ defined from $P$ by $a \le b :\iff (b - a \in P$ or $a = b)$ (Ordered field).

[A1]

Axiom (O1): for each $x \in F$ exactly one of $x \in P$ , $x = 0$ , $-x \in P$ holds (Ordered field).

[A2]

Axiom (O2): if $x, y \in P$ then $x + y \in P$ and $xy \in P$ (Ordered field).

[A3]

The order of an ordered field is defined by $a < b :\iff b - a \in P$ , and $a \le b$ means $a < b$ or $a = b$ (Ordered field).

[L1]

Every field is a commutative ring with $1 \ne 0$ (Every field is a commutative ring with $1 \ne 0$ ; it is an integral domain, and it is a commutative division ring, Field).

[L2]

In a ring, a subset satisfying trichotomy and closure under addition and multiplication is a cone, and the relation it induces is a total order making the ring an ordered ring whose positive cone is that subset (The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = \{\, x : 0 < x \,\}$ satisfies trichotomy and closure, and $a < b :\iff b - a \in P$ recovers the order, Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication).

[L3]

In an ordered field, $a \ne 0$ implies $a^{2} > 0$ , that is $a \cdot a \in P$ (Squares of nonzero elements are positive).

Proof

technique · direct

1.1

$F$ is a commutative ring under its own addition and multiplication, with the same $0$ and $1$ .

2.1

$P$ is a cone in the ring $F$ : trichotomy is axiom (O1) verbatim, and closure under addition and under multiplication is axiom (O2) verbatim. This proves claim 1.

step 1.1A1A2L2

3.1

By [L2] applied to the ring $F$ and the cone $P$ , the relation $a \le_P b :\iff (b - a \in P$ or $a = b)$ is a total order making $F$ an ordered ring, and its positive cone is $P$ .

step 2.1L2

4.1

That relation is the order of the ordered field: [A3] defines $a < b$ as $b - a \in P$ and $a \le b$ as $a < b$ or $a = b$ , which is the definition of $\le_P$ word for word. So $\le$ and $\le_P$ are the same relation, and claim 2 follows.

step 3.1A3

5.1

Claim 3: $1 \ne 0$ by [L1], so $1 \cdot 1 \in P$ by [L3]; and $1 \cdot 1 = 1$ because $1$ is the multiplicative identity. Hence $1 \in P$ , that is $0 < 1$ .

step 4.1L1L3∎

Remarks

This is one of the four bridges the page carries. The published Ordered field and the Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication of this page are two presentations of compatible orders, and without this item a later result about ordered rings could not be applied to $\mathbb{Q}$ or $\mathbb{R}$ without redoing the translation each time.
The converse is false as stated, and no converse is claimed. $\mathbb{Z}$ is an ordered ring and is not a field at all, as the companion page records. What is true, and is claim 2, is that on a field the two presentations of an order agree.
Where the strict form of (OR2) is used. Axiom (O2) of Ordered field is already strict, so claim 1 is a verbatim match with the cone conditions of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = \{\, x : 0 < x \,\}$ satisfies trichotomy and closure, and $a < b :\iff b - a \in P$ recovers the order and no strengthening is needed; that is exactly why Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication states its multiplicative axiom in the strict form.

Depends on

Used by

ℚ and ℝ are fields, hence commutative rings, integral domains and ordered rings, all of characteristic 0 Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Ordered field (Wikipedia) (standard reference, not scraped)
Ordered ring (Wikipedia) (standard reference, not scraped)
Field (mathematics) (Wikipedia) (standard reference, not scraped)